Image height for candle in front of converging mirror

[antiflag403]

Hey all,
Grade 11 physics assignment due tmrw and I am having some trouble. Any help would be appreciated. So here it is:

A candle of height 4.8 cm is placed 38.6 cm in front of a converging mirror with curvature 30.2 cm. What is the height of the image?

Heres what I have:

1/f = 1/Di + 1/ Do so... 1/15.1 = 1/Di + 1/38.6 so Di=2.56

then...

Hi/Ho = -Di/Do so... Hi/4.8 = -2.56/38.6 so Hi=-.32

where:
Ho= Height of object (candle)
Do= Distance of object
Di= Distance of image
Hi= Height of image
f=focus (which i have calculated to be 15.1)

The answer i get is -.32cm which doesn't seem right to me. Am i doing something wrong? Missing something??

As well... a totally unrelated question- When looking at interference patterns of water through 2 openings, if the second nodal line is formed at an angle of 50 degrees, will n=1 or n=1/2?

Thanks for any help!

 

[maverick280857]

Are you unsure of the sign or the magnitude of your answer? Or both?

 

[Chronos]

Try solving the image problem backwards.